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djverab [1.8K]
3 years ago
12

Which expression represents “seventeen more than one-fourth y”

Mathematics
2 answers:
vlabodo [156]3 years ago
5 0

Answer:

Expression which represents “seventeen more than one-fourth y” is:

\dfrac{1}{4}y+17

Step-by-step explanation:

We have to find a expression which represents

“seventeen more than one-fourth y”

one-fourth y= \dfrac{1}{4}y

So, seventeen more than one-fourth y= \dfrac{1}{4}y+17

Hence, expression which represents “seventeen more than one-fourth y” is:

\dfrac{1}{4}y+17

SpyIntel [72]3 years ago
3 0

hello again

17-1/49 is your correct answer thank you and have a great day please mark me as brainliest if can

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) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
Agustin solved an equation as shown. What error did Agustin make? What is the correct answer?
Hitman42 [59]

Subtract 3/4x from both sides

1/3x-4-3/4x= 3/4x+1-3/4x

-5/12x-4=1

Add 4 to both sides

-5/12x-4+4=1+4

-5/12x=5

Multiply both sides by 12/(-5)

(12/-5)*(-5/12x)=  (12/-5)*(5)

x= 12


I hope that's help.

3 0
3 years ago
Tomas wrote the equation y = 3x +3/4. When Sandra wrote her equation, they discovered that her equation had all the same solutio
mamaluj [8]
–6x + 2y = 3/2

or, 2y= 6x + 3/2
or, y= (6x + 3/2)/2
or, y = 6x/2 + 3/4
or, y = 3x + 3/4

Therefore, –6x + 2y = 3/2 is Sandra's equation.  


4 0
3 years ago
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What is the quotient of 3\4 and 5\6
Elis [28]

Answer:

9/10

Step-by-step explanation:

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Gloria Trench checked her gas mileage and found that she had used 16.6 gal of gas to travel 372 mi. At this rate, how many gallo
Andru [333]
About 127.156 its not exact but instead of 2850 miles i used 2849.52 miles so tell me if thats fine



5 0
3 years ago
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